Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 24.0{\rm m}/{\rm s} as it leaves the hose nozzle. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation of the hose until the water takes 3.30{\rm s} to reach a building 45.0{\rm m} away. You can ignore air resistance; assume that the end of the hose is at ground level.
You have omitted the question itself. See the first Related Question below for the full question.
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The 35 m of horizontal travel in 3.0 s tell you the horizontal velocity component, 11.67 m/s. The hose angle is the angle whose cosing is 11.67/27.0
At the highest point of the trajectory, the water's vertical velcoity component is zero. You should know what the horizintal component is.. it does not change.
To solve this problem, we can use the principles of projectile motion and apply them to the water shooting out of the hose. Let's break down the problem step by step:
Step 1: Analyzing the given information:
- Initial speed of water (v₀) = 24.0 m/s
- Time taken to reach the building (t) = 3.30 s
- Distance to the building (d) = 45.0 m
Step 2: Breaking down the motion:
Since the water is moving in projectile motion, we can separate its motion into horizontal and vertical components.
The horizontal motion:
- There is no horizontal force acting on the water, so its horizontal velocity remains constant throughout the motion.
- The horizontal distance traveled by the water (d_horizontal) is equal to the distance to the building (d).
The vertical motion:
- The only force acting on the water in the vertical direction is gravity.
- The initial vertical velocity of the water (v₀y) is determined by the angle of elevation of the hose.
- The time taken for the water to reach the maximum height (t_half) is half of the total time (t).
Step 3: Calculating horizontal motion:
Since there is no acceleration in the horizontal direction, we can use the formula:
d_horizontal = v₀ * t
Plugging in the values:
d_horizontal = 24.0 m/s * 3.30 s
d_horizontal = 79.2 m
So, the horizontal distance traveled by the water is 79.2 m.
Step 4: Calculating vertical motion:
Using the vertical motion information, we can calculate the initial vertical velocity (v₀y) of the water.
First, let's calculate the time taken for the water to reach the maximum height (t_half):
t_half = t / 2
t_half = 3.30 s / 2
t_half = 1.65 s
Now, we can use the formula to calculate the vertical displacement (d_vertical) using the time taken for half the total time:
d_vertical = v₀y * t_half + (1/2) * (-9.8 m/s²) * (t_half)²
Since the vertical displacement is zero at the maximum height (the water will return to the same height it started from), the above formula becomes:
0 = v₀y * t_half - (4.9 m/s²) * (t_half)²
Simplifying the equation:
0 = v₀y * (1.65 s) - (4.9 m/s²) * (1.65 s)²
Step 5: Solving for v₀y:
Now, we can solve the equation to find the initial vertical velocity (v₀y) of the water.
v₀y * (1.65 s) = (4.9 m/s²) * (1.65 s)²
Dividing both sides of the equation by (1.65 s):
v₀y = (4.9 m/s²) * (1.65 s)
v₀y = 8.08 m/s
So, the initial vertical velocity of the water is 8.08 m/s.
Step 6: Calculating the angle of elevation:
Now that we know the initial vertical velocity (v₀y) and the horizontal velocity (v₀), we can calculate the angle of elevation (θ) using trigonometry.
The tangent of the angle of elevation is given by:
tan(θ) = v₀y / v₀
Plugging in the values:
tan(θ) = 8.08 m/s / 24.0 m/s
Now, let's find the angle by taking the arctan of both sides:
θ = arctan(8.08 m/s / 24.0 m/s)
Using a calculator, we find that:
θ ≈ 19.7°
So, the angle of elevation of the hose is approximately 19.7°.